anonymous
  • anonymous
find the critical points: s(t)= (t-1)^4 (t+5)^3
Calculus1
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
first we find the derivative.. i know that.. then...?
anonymous
  • anonymous
then equate it to 0
jennychan12
  • jennychan12
set it = 0

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anonymous
  • anonymous
but wait, im confused to find the derivative for this one..
anonymous
  • anonymous
s'(t) =0
jennychan12
  • jennychan12
chain rule/product rule
anonymous
  • anonymous
its 4(t-1)^3 3(t+5)^2
anonymous
  • anonymous
Yup
anonymous
  • anonymous
what's next after it? i'm a little lost after it
anonymous
  • anonymous
4(t-1)^3 3(t+5)^2 = 0
anonymous
  • anonymous
Now Find t
anonymous
  • anonymous
what about chain rule?
jennychan12
  • jennychan12
doesnt apply here because derviative of t is just 1
anonymous
  • anonymous
if ur function is of the type (t^2 + 2)^4 then s'(t) = 4 (t^2+2) * 2t
anonymous
  • anonymous
sorry s'(t) = 4 (t^2+2)^3 * 2t
anonymous
  • anonymous
oh right.
anonymous
  • anonymous
so if the derivative of t is 1 then how do i solve t?
anonymous
  • anonymous
sorry imm not sure what to do after ... 4(t-1)^3 3(t+5)^2

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